Question

The percentage dissociation of a 0.011 m aqueous solution of $K_3[Fe(CN{)}_6]$ which freezes at $-0.063^{\circ }C$ is
($K_f$ for water $=1.86$)

• A. 75%
• B. 67%
• C. 33%
• D. 50%

Answer 1

Answer: (B)

$\Delta T_f=K_f\times m=1.86\times 0.011=0.021$
$\Delta T_f$ (calculated) $=0.021{}^{\circ }C$
$\Delta T_f$ (observed) $=0.063{}^{\circ }C$ (given)
$i=\frac{Observed\Delta T_f}{Calculated\Delta T_f}=\frac{0.063}{0.021}=3$
Degree of dissociation, $\alpha =\frac{i-1}{n-1}$
$K_3\left[Fe{\left(CN\right)}_6\right]\rightleftharpoons 3K^{+}+{\left[Fe\left(C{N}_6\right)\right]}^{3-}$ thus, $n=4$
$\alpha =\frac{3-1}{4-1}=0.67\Rightarrow$ Degree of dissociation $=67\%$